Author Topic: M1 doubt  (Read 905 times)

Offline sweetest angel

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M1 doubt
« on: December 16, 2009, 05:54:15 am »
june 2009 Q2 b) (attached)

this is the answer

R = (i - 3j) + (pi + 2pj) = (1 + p)i + (-3 + 2p)j
R is parallel to i ? (-3 + 2p) = 0
? p = 3/2

but i dnt get it. if R is parallel to i then the value of i=0 not j=0 ?
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Offline T.Q

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Re: M1 doubt
« Reply #1 on: December 16, 2009, 03:22:35 pm »
june 2009 Q2 b) (attached)

this is the answer

R = (i - 3j) + (pi + 2pj) = (1 + p)i + (-3 + 2p)j
R is parallel to i ? (-3 + 2p) = 0
? p = 3/2

but i dnt get it. if R is parallel to i then the value of i=0 not j=0 ?

because i  is a straight horizantal line so it will have a zero j
and because R is parallel to i ,then R will have the value of j equal to zero
thats why they took the j value equal to zero
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Offline sweetest angel

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Re: M1 doubt
« Reply #2 on: December 16, 2009, 04:40:13 pm »
ohhhhh...but we dnt knw if it passes through j or not so maybe it has a j value  ???
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Offline eightAs

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Re: M1 doubt
« Reply #3 on: December 25, 2009, 08:31:28 am »
ohhhhh...but we dnt knw if it passes through j or not so maybe it has a j value  ???
consider the whole set up on the  xy - plane
consider an arbitrary vector and the unit vector \vec{\imath}
We assume both vectors are parallel.
Lets assume the arbitrary vector is a \vec{\imath}+ b \vec{\jmath}
Gradient of \vec{\imath} = Gradient of the other vector
 \frac{\Delta y1}{\Delta x1} = \frac{\Delta y2}{\Delta x2}<br />\frac{0}{1} = \frac{b}{a}<br />\Rightarrow b = 0.<br />
Thus if an arbitrary vector and \vec{\imath} are parallel, the co-effecient of j for the arbitrary vector has to equal 0.
« Last Edit: December 30, 2009, 02:05:11 pm by eightAs »
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Offline Ghost Of Highbury

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Re: M1 doubt
« Reply #4 on: December 30, 2009, 10:20:37 am »
ohhhhh...but we dnt knw if it passes through j or not so maybe it has a j value  ???

a vector (2D) has two components; i and j which define its direction.
It cannot "PASS" through j. its a component. It can pass through a point.

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